I meant to write a longer blog entry on strong measure zero sets (on the real line ), but it is getting too long, so it may take me more than I expected. For now, let me record here an argument showing the following:

Theorem. If is a strong measure zero set and is a closed measure zero set, then has measure zero.

where the result is shown for strong measure zero subsets of . This is actually the easy direction of Pawlikowski’s result, showing that this condition actually characterizes strong measure zero sets, that is, if is measure zero for all closed measure zero sets , then is strong measure zero. (Since this was intended for my analysis course, and I do not see how to prove Pawlikowski’s argument without some appeal to results in measure theory, I am only including here the easy direction.) Pawlikowski’s argument actually generalizes an earlier key result of Galvin, Mycielski, and Solovay, who proved that a set has strong measure zero iff it can be made disjoint from any given meager set by translation, that is, iff for any meager there is a real with disjoint from .

For the third talk (and a link to the second one), see here. The fourth talk took place on October 12.

We want to show the following version of Theorem 2:

Theorem.Suppose is a singular strong limit cardinal of uncountable cofinality. Then the following are equivalent:

For each ideal on , player II has a winning coding strategy in .

.

Since has uncountable cofinality, option 2 above is equivalent to saying that the instance of corresponding to holds.

Before we begin the proof, we need to single out some elementary consequences in cardinal arithmetic of the assumptions on . First of all, since is singular strong limit, then for any cardinal , we have that

Also, since the cofinality of is uncountable, we have Hausdoff’s result that if , then . I have addressed both these computations in my lecture notes for Topics in Set Theory, see here and here.

We are ready to address the Theorem.

Proof. We use Theorem 1. If option 1. fails, then there is an ideal on with .

Note that , and . Moreover, if , then since, otherwise,

.

So and then, by Hausdorff, in fact , and option 2. fails.

Suppose option 2. fails and let , so and . We use to build an ideal on with .

For this, we use that there is a large almost disjoint family of functions from into . Specifically:

Lemma. If is singular strong limit, there is a family with and such that for all distinct , we have that .

In my notes, I have a proof of a general version of this result, due to Galvin and Hajnal, see Lemma 12 here; essentially, we list all functions , and then replace them with (appropriate codes for) the branches they determine through the tree . Distinct branches eventually diverge, and this translates into the corresponding functions being almost disjoint.

Pick a family as in the lemma, and let be a subfamily of size . Let . We proceed to show that and use to define an ideal on as required.

First, obviously . Since and , it follows that , or else , since is strong limit.

Now define

Clearly, is an ideal. We claim that . First, each singleton with is in , so . Define by . Since the functions in are almost disjoint, it follows that is 1-1. Let be the image of . By construction, is cofinal in . But then

,

where the first inequality follows from noticing that any has size at most . It follows that , as claimed.

Finally, we argue that , which completes the proof. For this, consider a cofinal , and a map such that for all , we have .

Since is cofinal in , it follows that is cofinal in . But this gives the result, because

In the last Corollary of the Appendix to lecture I.5, I indicate that in we have that

whenever is not for some infinite limit ordinal In fact,

holds.

This result is best possible in terms of positive results. In Theorem 11 of the paper by John Hickman listed at the end, it is shown that for any such it is consistent with that there is an with for which

I also want to give an update on the topics discussed in lecture III.3.

and Hajnal asked whether it is possible to have infinite cardinals such that

Galvin and Prikry showed (see Corollaries 16 and 18 of lecture III.3) that any such must be larger than and that

Following a nice suggestion of Grigor Sargsyan, we use arguments as in Theorem 9 from lecture III.5 to show that this partition relation cannot hold.

The key is the following:

Lemma 1 If there are infinite cardinals such that then for every sufficiently large there is an elementary embedding such that and

Here is a brief sketch:

Proof: By Corollary 20 from lecture III.3, the given relation is equivalent to Consider a -Skolem function so that any closed under is both closed under -sequences and an elementary substructure of

Use to define a coloring by setting whenever and otherwise. By assumption, there is with Note that if is the closure of under then But we can assure that and the result follows by taking as the transitive collapse of

One concludes the proof by noting that it is impossible to have such embeddings. For this, it suffices that and that admits a fixed point past its critical point. One then obtains a contradiction just as in Kunen’s proof that there are no embeddings see Corollary 9 in lecture III.3.

Similarly, Matthew Foreman has shown that there are no embeddings with closed under -sequences. The reason is that any such embedding must admit a fixed point past its critical point, as can be argued from the existence of scales. See the paper by Vickers and Welch listed at the end for a proof of this result.

On the other hand, it is still open whether one can have embeddings such that computes cofinality correctly.

The only reference I know for precisely these matters is the handbook chapter MR2768702. Koellner, Peter; Woodin, W. Hugh. Large cardinals from determinacy. In Handbook of set theory. Vols. 1, 2, 3, 1951–2119, Springer, Dordrecht, 2010. (Particularly, section 7.) For closely related topics, see also the work of Yong Cheng (and of Cheng and Schindler) on Harr […]

As other answers point out, yes, one needs choice. The popular/natural examples of models of ZF+DC where all sets of reals are measurable are models of determinacy, and Solovay's model. They are related in deep ways, actually, through large cardinals. (Under enough large cardinals, $L({\mathbb R})$ of $V$ is a model of determinacy and (something stronge […]

Throughout the question, we only consider primes of the form $3k+1$. A reference for cubic reciprocity is Ireland & Rosen's A Classical Introduction to Modern Number Theory. How can I count the relative density of those $p$ (of the form $3k+1$) such that the equation $2=3x^3$ has no solutions modulo $p$? Really, even pointers on how to say anything […]

(1) Patrick Dehornoy gave a nice talk at the Séminaire Bourbaki explaining Hugh Woodin's approach. It omits many technical details, so you may want to look at it before looking again at the Notices papers. I think looking at those slides and then at the Notices articles gives a reasonable picture of what the approach is and what kind of problems remain […]

It is not possible to provide an explicit expression for a non-linear solution. The reason is that (it is a folklore result that) an additive $f:{\mathbb R}\to{\mathbb R}$ is linear iff it is measurable. (This result can be found in a variety of places, it is a standard exercise in measure theory books. As of this writing, there is a short proof here (Intern […]

The usual definition of a series of nonnegative terms is as the supremum of the sums over finite subsets of the index set, $$\sum_{i\in I} x_i=\sup\biggl\{\sum_{j\in J}x_j:J\subseteq I\mbox{ is finite}\biggr\}.$$ (Note this definition does not quite work in general for series of positive and negative terms.) The point then is that is $a< x

The result was proved by Kenneth J. Falconer. The reference is MR0629593 (82m:05031). Falconer, K. J. The realization of distances in measurable subsets covering $R^n$. J. Combin. Theory Ser. A 31 (1981), no. 2, 184–189. The argument is relatively simple, you need a decent understanding of the Lebesgue density theorem, and some basic properties of Lebesgue m […]

Given a class $S$, to say that it can be proper means that it is consistent (with the axioms under consideration) that $S$ is a proper class, that is, there is a model $M$ of these axioms such that the interpretation $S^M$ of $S$ in $M$ is a proper class in the sense of $M$. It does not mean that $S$ is always a proper class. In fact, it could also be consis […]

As the other answers point out, the question is imprecise because of its use of the undefined notion of "the standard model" of set theory. Indeed, if I were to encounter this phrase, I would think of two possible interpretations: The author actually meant "the minimal standard model of set theory", that is, $L_\Omega$ where $\Omega$ is e […]